Stress transformation under axial tension: A member of normal cross-section A is subjected to a tensile force P. What is the normal stress on an oblique plane inclined at angle θ to the transverse (normal) plane?

Introduction / Context:Determining stresses on inclined planes helps predict failure planes and understand combined normal and shear actions even when the external load is uniaxial.

Given Data / Assumptions:

• Tension P on a uniform member with area A.
• Angle θ is measured from the transverse cross-section.
• Linear elasticity and small deformations.

Concept / Approach:Under axial tension, the traction on an inclined plane has magnitude (P/A) * cos θ and decomposes into normal and shear components. Standard results: σ_n = (P/A) * cos^2 θ and τ = (P/A) * sin θ * cos θ.

Step-by-Step Solution:Find inclined area A_inclined = A / cos θ.Traction magnitude p = P / A_inclined = (P/A) * cos θ.Normal component σ_n = p * cos θ = (P/A) * cos^2 θ.

Verification / Alternative check:Check limits: θ = 0° → σ_n = P/A; θ = 90° → σ_n = 0, consistent with mechanics.

Why Other Options Are Wrong:sin^2 θ form swaps boundary values; shear expression sin θ cos θ is not a normal stress; sec^2 θ and zero are not generally correct.

Common Pitfalls:Confusing angle definition or mixing up σ_n and τ.

Final Answer:(P / A) * cos^2 θ.